How the fuck do you prove this without truth tables? algebraically isn't this as simple as it gets?

>t. engineer who took "Intro to Logic For Engineers"

No I don't, q == ~~q

Truth table method is actually faster than any of the proofs listed here, even if you count by character length.

Informal proofs

informal
informal and wrong

It's a 100% rigorous proof acceptable in any mathematical logic course. Only high schoolers do proofs by pattern matching axioms.

Truth tables blow up exponentially and have 0 insight to them. You should never look at them beyond day 1 freshman year.

>t. state school engineer

>samefagging butthurt CS major

go to bed

op here. i can't believe i missed that definition of the biconditional. i thought i could only use conjuctive and disjunctive clauses.

The is that way, go home brainlet.

here are the other ones in case you were interested. im gonna try them out using that [(p=>q ^ (q=>p)] format of the definition rather than the conjunction.

What kind of autism do you have that you think calling someone a CS major is an insult?

I'm a statistics grad student fyi

>grad student
>can't into basic logic

Sure you are, sure.

where am i wrong

You're the one claiming is wrong and are spamming the thread multiple times at that.

And its not a proper proof. Where am I wrong?

P IFF Q === P IMPLIES Q AND Q IMPLIES P
P IMPLIES Q AND Q IMPLIES P ===
(NOT P OR Q) AND (NOT Q OR P)

It's a 100 percent rigorous proof. You're not articulating any counter argument.

>Thinking physics is mathematical
>Thinking DEs are complicated
>Rock me kekadeus

I never said it wasn't rigorous.
Its informal. There is a difference, but I guess only "CS brainlets" know this.

OP here there is no way to simplify 18 if it's just a conditional is there? maybe (p v -q) but how would that lead back to the implication?

still wrong.

How do you know how many IPs are in this thread?

You said twice that it was wrong: and Protip: you can't hide behind anonymity if there are only 3 IPs in the thread (op, you, and I)

>hurr durr, you're not using 2 column proofs like philosophy 101 students do so it must be wrong.

~(pq)
~(p->q & q->p) // definition
~(p->q) or ~(q->p) //De Morgan
~(~p or q) or ~(~q or p) //definition
(p & ~q) or (q & ~p) // De Morgan
(p ~q) //by ex16

See, not wrong anywhere.

>informal

No one uses "formal" proofs like it's defined in mathematical logic to actually prove stuff. It's only used to prove theorems on metamathematics and with ATP.

>so called graduate student in stats
>thinks calc 2's DEs is the highest math in physics

Stop lying on the internet. Data ""science"" isn't a statistics degree.

>being this new

>still wrong

Just deal with yourself

All of these kinds of problems are solved by unpacking the definitions and meeting in the middle

(p->q) = ~p or q
(~q->~p) = ~~q or ~p = q or ~p = ~p or q

QED

Why is this man's penis so small?

>not an argument

>Ad hominem

let me guess... CS Major?

...

is it just me or is this guys wiener oddly small

...

...

>all those (you)'s

...

so i can't go from one to the other? damn this doesn't seem hard but for some reason i keep fudging up the steps when i see no equivalence blatantly in my face when unpacking from the implication to the negation.

1(p&q)v(~p&~q)
2||p
case 1
3|||p&q velim1
4|||q &elim3
case 2
5|||~p&~q velim1
6||| ~p &elim5
7|||q cont intro 2,6
8|| p->q ->intro 2,7
9|| q
case 1
10|||p&q velim1
11|||p &elim10
case 2
12|||~p&~q velim1
13||| ~q &elim12
14|||p cont intro 9,13
15||q->p ->intro 9,14
16|pq intro 9,15

i think this is sort of what you want it to look like but the other way around
just being like "HUR DUR THIS IS A DEFINITION" probably isn't the right answer
but what the fuck do i know im an engineer so im gonna go suck some robot dicks
also i havent done this shit in forever so i probably forgot the correct use of velim or &elim since i made them do the same thing even though i was just trying to prove by cases

last line should be
16|pq intro 8,15

...

>so i can't go from one to the other

No, you can. I'm just saying it's easier to meet in the middle rather than try to figure out how to repack it directly.

>Question
Show statement A == statement Z

>Scrap Paper
A == B == C ... == R == S
Z == Y == X ... == T == S

>Actual proof
A == B == C ... == R == S == T ... == X == Y == Z

ok i think i get what you mean. i just did exercises 20 and got this.

prove: -(p xor q) = (pq)

p xor q = (p v q) ^ (-p ^ -q)

p q = (p=>q) ^ (q=>p)
=(-p v q) ^ (-q v p)

- ( p xor q) = - [(p v q) ^ (-p v -q)]
=[(-p ^ -q) v (p ^ q)]
=pq

QED? i did De Morgan's Law on the -p v -q...

is it proper form to write a proof as two things being equal to the same thing, therefore they're equal to each other?

Euclid's first axiom:

Things which are equal to the same thing are equal to each other.

is there a proof for this?

if i have

(-p^q) v (-p^r) v (-q^r) v (r)

can i eliminate (-p^r) v (-q^r) to just have (-p^q) v (r)? this is for 23 from

Replace the equals sign and prove the tautology

Correct proofs, thanks for posting

samefag

kys

[math] p \iff q \\
(p \implies q) \land (q \implies p) \text{ Definition of equivalence} \\
( \neg p \lor q) \land ( \neg q \lor p) \text{ Definition of implication} \\
(( \neg p \lor q) \land \neg q) \lor (( \neg p \lor q) \land p) \text{ Distribution } \\
((\neg q \land q) \lor (\neg q \land \neg p)) \lor ((p \land \neg p) \lor (p \land q)) \text{ Distribution} \\
( \neg q \land \neg p) \lor (p \land q) \text{ Removal of contradictions} \\
(p \land q) \lor ( \neg p \land \neg q) \text{ Commutativity } [/math]

God fucking damn. I did propositional logic as a freshman and I was never ever ever asked to prove this. Holy fuck your professor is savage but it can be done.

Note that I haven't really studied more propositional logic after that so if I could do it you should be able to do it.

Yes, just factor out
(-p&r) or (-q&r) or (r) = (-p or -q or 1)&r = 1&r = r

>being unable to tell who's who

ip 1 (OP)
ip 2 (faggot)
ip 3 (me)
ip 4
ip 5
ip 6
ip 7
ip 8
ip 9
ip 10

how does that work???

Did u try distributing?

Boolean algebra. Let * be AND and + be OR with AND being higher in precedence.

A*(B+C)=A*B+A*C //Distributivity
A+(B*C)=(A+B)*(A+C) //looks weird but follows from DeMorgan'ing the 1st one
A+1 = 1
A+0 = A
A*1 = A
A*0 = 0

but how can you ignore the p and q?

Nice photoshop on the picture bro. Underline @ the last 9 was cut way too early. But nice try bro

-p or -q or true
-p or (-q or true)
-p or (true)
true

you get a T from factoring out r though? what definition lets you do that?

r = true & r

You want to prove a semantic equivalency, which means you must show a semantic arrow going both ways. The left side's equivalency is not semantic btw, it's syntactic.

Are you in my summer class? A school in NC?

[math]p\leftrightarrow q = p\rightarrow q \wedge p\leftarrow q [/math]
[math]\qquad \quad = (\neg p \vee q)\wedge(p\vee\neg q) [/math]
[math]\qquad \quad = (p \wedge \neg p) \vee (\neg p \wedge \neg q) \vee (p \wedge q) \vee (\neg q \wedge q)[/math]
[math]\qquad \quad = (p \wedge q) \vee (\neg p \wedge \neg q)[/math]

You're half done since you've only shown [math]\Leftarrow[/math]

Next you must show [math]\Rightarrow[/math]

First we seek to show the forward direction

[math]1 (1) p \iff q, \ \text{Assumption}[/math]
[math]1 (2) (p \Rightarrow q) \land (q \Rightarrow p), \ \text{Definition of the biconditional} [/math]
[math]1 (3) p \Rightarrow q, \ 2 \ \text{Conjunction elimination} [/math]
[math]1 (4) q \Rightarrow p, \ 2 \ \text{Conjunction elimination}[/math]
[math](5) p \lor \lnot p , \ \text{Theorem Introduction: Law of excluded middle} [/math]
[math]6 (6) p, \ \text{Assumption} [/math]
[math]1, 6 (7)q, \ 6,3 \ \text{Modus ponens} [/math]
[math]1, 6 (8) p \land q, \ 6,7 \ \text{Conjunction introduction} [/math]
[math]1, 6 (9) (p \land q) \lor (\lnot p \land \lnot q) , \ 8 \ \text{Disjunction introduction} [/math]
[math]10 (10) \lnot p, \ \text{Assumption} [/math]
[math]1, 10 (11) \lnot q, \ 4,10 \ \text{Modus tolens}[/math]
[math]1, 10 (12) \lnot p \land \lnot q, \ 10,11 \ \text{Conjunction introduction} [/math]
[math]1, 10 (13) (p \land q) \lor (\lnot p \land \lnot q), \ 12 \ \text{Disjunction introduction} [/math]
[math]1 (14) (p \land q) \lor (\lnot p \land \lnot q), \ 5, 6, 9, 10, 13, \ \text{Disjunction elimination} [/math]

The other direction is an exercise

equality is symmetric

You have to prove both sides retard

Can't you read from right to left, brainlet?

Oh, Rosen's discrete maths, we meet again.